4b: probability part b normal distributions
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4B: Probability part B Normal Distributions. How’s my hair?. Looks good. The Normal distributions. Last lecture covered the most popular type of discrete random variable: binomial variables This lecture covers the most popular continuous random variable: Normal variables - PowerPoint PPT PresentationTRANSCRIPT
HS 167 4B: Probability Part B 1
4B: Probability part BNormal Distributions
HS 167 4B: Probability Part B 2
The Normal distributions
Last lecture covered the most popular type of discrete random variable: binomial variables This lecture covers the most popular continuous random variable: Normal variables History of the Normal function
Recognized by de Moivre (1667–1754)
Extended by Laplace (1749–1827)
How’s my hair?
Looks good.
HS 167 4B: Probability Part B 3
Probability density function (curve)Illustrative example: vocabulary scores of 947 seventh gradersSmooth curve drawn over histogram is a density function model of the actual distributionThis is the Normal probability density function (pdf)
HS 167 4B: Probability Part B 4
Areas under curve (cont.)
Last week we introduced the idea of the area under the curve (AUC); the same principals applies hereThe darker bars in the figure represent scores ≤ 6.0, About 30% of the scores were less than or equal to 6Therefore, selecting a score at random will have probability Pr(X ≤ 6) ≈ 0.30
HS 167 4B: Probability Part B 5
Areas under curve (cont.)Now translate this to a
Normal curveAs before, the area under
the curve (AUC) = probability
The scale of the Y-axis is adjusted so the total AUC = 1
The AUC to the left of 6.0 in the figure to the right (shaded) = 0.30
Therefore, Pr(X ≤ 6) ≈ 0.30In practice, the Normal
density curve helps us work with Normal probabilities
HS 167 4B: Probability Part B 6
Density Curves
HS 167 4B: Probability Part B 7
Normal distributionsNormal distributions = a family of distributions with
common characteristics Normal distributions have two parameters
Mean µ locates center of the curve Standard deviation quantifies spread (at points of
inflection)
Arrows indicate points of inflection
HS 167 4B: Probability Part B 8
68-95-99.7 rule for Normal RVs
68% of AUC falls within 1 standard deviation of the mean (µ )
95% fall within 2 (µ 2)
99.7% fall within 3 (µ 3)
HS 167 4B: Probability Part B 9
Illustrative example: WAISWechsler adult intelligence scores (WAIS)
vary according to a Normal distribution with μ = 100 and σ = 15
HS 167 4B: Probability Part B 10
Illustrative example: male height
Adult male height is approximately
Normal with µ = 70.0 inches and
= 2.8 inches (NHANES, 1980)
Shorthand: X ~ N(70, 2.8)
Therefore: 68% of heights = µ = 70.0 2.8 = 67.2 to 72.8 95% of heights = µ 2 = 70.0 2(2.8) = 64.4 to
75.6 99.7% of heights = µ 3 = 70.0 3(2.8) = 61.6 to
78.4
HS 167 4B: Probability Part B 11
Illustrative example: male height
What proportion of men are less than 72.8 inches tall? (Note: 72.8 is one σ above μ)
?
70 72.8 (height)
+1
84%
68% (by 68-95-99.7 Rule)
16%
-1
16%
HS 167 4B: Probability Part B 12
Male Height Example
What proportion of men are less than 68 inches tall?
?
68 70 (height)
68 does not fall on a ±σ marker.To determine the AUC, we must first standardize
the value.
HS 167 4B: Probability Part B 13
Standardized value = z score
x
z
To standardize a value, simply subtract μ and divide by σ
This is now a z-score
The z-score tells you the number of standard deviations the value falls from μ
HS 167 4B: Probability Part B 14
Example: Standardize a male height of 68”
71.08.2
7068
x
z
Recall X ~ N(70,2.8)
Therefore, the value 68 is 0.71 standard deviations below the mean of the distribution
HS 167 4B: Probability Part B 15
Men’s Height (NHANES, 1980)
-0.71 0 (standardized values)68 70 (height values)
?
What proportion of men are less than 68 inches tall? = What proportion of a Standard z curve is less than –0.71?
You can now look up the AUC in a Standard Normal “Z” table.
HS 167 4B: Probability Part B 16
Using the Standard Normal table
z .00 .01 .02
0.8 .2119 .2090 .2061
0.7 .2420 .2389 .2358
0.6 .2743 .2709 .2676
Pr(Z ≤ −0.71) = .2389
HS 167 4B: Probability Part B 17
Summary (finding Normal probabilities)
Draw curve w/ landmarksShade areaStandardize value(s)Use Z table to find appropriate AUC
-0.71 0 (standardized values)68 70 (height values)
.2389
HS 167 4B: Probability Part B 18
Right tail
What proportion of men are greater than 68” tall?Greater than look at right “tail”Area in right tail = 1 – (area in left tail)
-0.71 0 (standardized values)68 70 (height values)
.2389 1- .2389 = .7611
Therefore, 76.11% of men are greater than 68 inches tall.
HS 167 4B: Probability Part B 19
Z percentiles • zp the z score with
cumulative probability p
• What is the 50th percentile on Z? ANS: z.5 = 0
• What is the 2.5th percentile on Z? ANS: z.025 = 2
• What is the 97.5th percentile on Z? ANS: z.975 = 2
HS 167 4B: Probability Part B 20
Finding Z percentile in the table
Look up the closest entry in the table
Find corresponding z score
e.g., What is the 1st percentile on Z?
z.01 = -2.33
closest cumulative proportion is .0099
z .02 .03 .04
2.3 .0102 .0099 .0096
HS 167 4B: Probability Part B 21
Unstandardizing a value
.10 ? 70 (height values)
How tall must a man be to place in the lower 10% for men aged 18 to 24?
HS 167 4B: Probability Part B 22
Use Table A
Look up the closest proportion in the table
Find corresponding standardized score
Solve for X (“un-standardize score”)
Table A:Standard Normal Table
HS 167 4B: Probability Part B 23
Table A:Standard Normal Proportion
z .07 .09
1.3 .0853 .0838 .0823
.1020 .0985
1.1 .1210 .1190 .1170
1.2
.08
.1003
Pr(Z < -1.28) = .1003
HS 167 4B: Probability Part B 24
Men’s Height Example (NHANES, 1980)
How tall must a man be to place in the lower 10% for men aged 18 to 24?
-1.28 0 (standardized values)
.10 ? 70 (height values)
HS 167 4B: Probability Part B 25
Observed Value for a Standardized Score
“Unstandardize” z-score to find associated x :
zx
zx
HS 167 4B: Probability Part B 26
Observed Value for a Standardized Score
x = μ + zσ = 70 + (1.28 )(2.8) = 70 + (3.58) = 66.42
A man would have to be approximately 66.42 inches tall or less to place in the lower 10% of the population